Class 9 NCERT Mathematics Solutions - Class IX CBSE Board Maths Questions | Lines and Angles - Exercise 6.3 Answers

CBSE Class IX NCERT Mathematics Solutions

Chapter 6 - Lines and Angles

IXth NCERT Mathematics Textbook Exercise 6.3 Solved

(Page 107, 108)

Q1: In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135º and ∠PQT = 110º, find ∠PRQ.

 Ans:  In ∆ PQR side QP and RQ are produced to point S and T respectively such that / SPR = 135^O, / PQT = 110^O. We have to find / PRQ.

As / SPR + / RPQ = 180^O (linear pair of angles).

Or, 135^O + / RPQ = 180^O

Or, / RPQ = 180^O – 135^O = 45^O

Now, / TPQ = / QPR + / PRQ

Or, 110^O = 45^O + / PRQ

Or, / PRQ = 65^O  

Q2: In the given figure, ∠X = 62º, ∠XYZ = 54º. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.

Ans:

As the sum of all interior angles of a triangle is 180º, therefore, for ΔXYZ,

∠X + ∠Y + ∠Z = 180º

Or, 62º + 54º + ∠Z = 180º

Or, ∠Z = 180º − 116º = 64^O

Now, ∠OZY = 64^O/2 = 32º (as OZ is the angle bisector of ∠Z)

Similarly, ∠OYZ = 54^O/2 = 27^O

Consider ΔOYZ,

∠OYZ + ∠YOZ + ∠OZY = 180º

Or, 27º + ∠YOZ + 32º = 180º

Or, ∠YOZ = 180º − 59º = 121º

Q3: In the given figure, if AB || DE, ∠BAC = 35º and ∠CDE = 53º, find ∠DCE.

Ans:

/ DEC = / BAC = 35^O ………. (i) [Alternate interior angles]

/ CDE = 53^O ………. (ii) [given]

In ∆CDE using angle sum property we have,

/ CDE + / DEC + / DCE = 180^O  

Or, 53^O + 35^O + / DCE = 180^O

Or, / DCE = 92^O

Q4: In the given figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40º, ∠RPT = 95º and ∠TSQ = 75º, find ∠SQT.

Ans: Applying angle sum property in ∆PRT we have,

/ PTR + / PRT + / RPT = 180^O

Or, / PTR + 40^O + 95^O = 180^O

Or, / PTR = 45^O

Or, / QTS = / PTR = 45^O [vertically opposite angles]

Applying angle sum property in ∆TSQ we have,

/ QTS + / TSQ + / SQT = 180^O

Or, / SQT + 45^O + 75^O = 180^O

Or, / SQT = 60^O

Q5: In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28º and ∠QRT = 65º, then find the values of x and y.

Ans:

/ QRT = / RQS + / QSR [as the exterior angle is equal to the sum of the two interior opposite angles].

Or, 65^O = 28^O + / QSR

Or,  / QSR = 37^O

Given that PQ ⊥ PS i.e. / QPS = 90^O

Or, PQ || SR.

So, / QPS + / PSR = 180^o [sum of the consecutive interior angles on the same side of the traversal is 180^O].

Therefore, 90^O + / PSR = 180^O

Or, / PSR = 90^O

Or, / PSR + / QSR = 90^O

Or, y + / QSR = 90^O

Or, y + 37^O = 90^O

Or, y = 53^O

Now consider ∆PQS,

/ PQS + / QSP + / QPS = 180^O

Or, x + 53^O + 90^O = 180^O

Or, x = 37^O    

Q6: In the given figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR= ½ ∠QPR.

Ans:

/ QTR = ½ / QPR

As RT and QT are the bisectors of / PRS and / PQR respectively,

Therefore, / PRT = / TRS and / PQT = / TQR.

=> / TRS = / T + / TQR

=> 2/ TRS = 2/ T + 2/ TQR

=> / PRS = 2/ T + / PQR

=> / P + / PQR = 2/ T + / PQR

Or, / P = 2/ T

Or, / T = / P/2

[Exterior angle = Sum of opposite interior angle].